Question

Which of the following transformation reduces the differential equation $\frac{dz}{dx}+\frac{z}{x}lnz=\frac{z}{x^2}(lnz{)}^2$ into the form $\frac{du}{dx}+p\left(x\right)u=Q\left(x\right)?$
Where P(x) and Q(x) are function of x.

• A. $u=lnz$
• B. $u=e^z$
• C. $u=(lnz{)}^{-1}$
• D. $u=(lnz{)}^2$

Answer 1

The correct option is C $u=(lnz{)}^{-1}$

$\frac{dz}{dx}+\frac{z}{x}lnz=\frac{z}{x^2}(lnz{)}^2$

dividing $e{q}^n$ by $z(lnz{)}^2$

$\frac{1}{z(lnz{)}^2}\frac{dz}{dx}+\frac{1}{xlnz}=\frac{1}{x^2}$

put $\frac{1}{lnz}=u$

$-\frac{1}{z(lnz{)}^2}\frac{dz}{dx}=\frac{du}{dx}$

$\therefore -\frac{du}{dx}+\frac{u}{x}=\frac{1}{x^2}$

$\frac{du}{dx}+P\left(x\right)u=Q\left(x\right)$

where $P\left(x\right)=\frac{-1}{x}$ and $Q\left(x\right)=\frac{1}{-x^2}$